One algebraic way to motivate this is to observe that the signs in the differential for the Hom are precisely what is needed for 0-cycles in the $\hom(A,B)$ complex to be the set of morphisms of complexes $A\to B$ (and also, that the 0th homology group $H_0(\hom(A,B))$ is the set of homotopy classes of morphisms $A\to B$). This is quite great.
Once you decide you want this, all the other signs you mention follow because you need various things to hold. For example, you want the adjuntion between $\hom$ and $\otimes$ to hold for the internal versions, so this forces you to add signs to the $\otimes$, and so on.
A topological, indirect explanation for the appearence of most signs is that in the long exact sequence of bases maps corresponding to a map $f:X\to Y$, which looks like
$$X\to Y\to C(f)\to S(X) \to S(Y) \to S(C(f)) \to \cdots$$ there is a "sign" which you cannot get rid of. This sign reproduces itself in every algebraic version.
Apart from the two factorization axioms, the proof in Dwyer-Spalinski's section 7 goes through in the desired generality and it doesn't make a detour via Dold-Kan. Therefore I'll only fill in the part of the argument depending on the small object argument in all the sources I know.
In order to produce factorizations without the small object argument, you can make use of the mapping cone and mapping cylinder constructions. I did this in my thesis, Appendix C, for bounded below complexes in an exact category with enough projectives. For nonnegative complexes, essentially the same constructions work, but one faces the additional difficulty of making sure the complexes stay non-negative. I'm not aware that such an elementary proof is written up anywhere in the literature, but it is not too difficult...
The argument is a rather tedious exercise in homological algebra. The explicit nature of the construction shows that you can obtain functorial factorizations whenever there is a 'projective resolvent functor', that is to say, there is a functor $P \colon \mathcal{A} \to \mathcal{P}$ and an epimorphic natural transformation $P \Rightarrow \operatorname{id}_{\mathcal{A}}$.
We will use the following lemma:
For every complex $A$ in $\mathcal{M}$ there is a degreewise epimorphic quasi-isomorphism $P \to A$, where $P$ is a complex with projective components.
A proof and references can be found here: Every chain complex is quasi-isomorphic to a $\mathcal J$-complex
Consider an arbitrary morphism $f \colon A \to B$ $\require{AMScd}$. We have to produce the usual factorizations.
Let's start with the easier case:
Recall that the usual map $\operatorname{cyl}(f) \to B$ is degreewise surjective and a chain homotopy equivalence, in particular it is both a fibration and a weak equivalence. The inclusion $A \to \operatorname{cyl}(f)$ is close to being a cofibration, but its cokernel $\operatorname{cone}(f)$ isn't a complex with projective components.
To remedy this, choose a degreewise epimorphic quasi-isomorphism $p\colon P \to \operatorname{cone}(f)$ from a complex with projective components $P$, and let $q$ be the pullback of $p$ along the projection $\operatorname{cyl}(f) \to \operatorname{cone}(f)$. We obtain the following commutative diagram:
$$
\begin{CD}
A @>i>> Q @>>> P \\
@| @V{\sim}VqV @V{\sim}VpV \\
A @>>> \operatorname{cyl}(f) @>>> \operatorname{cone}(f) \\
@| @V{\simeq}VV \\
A @>f>> B
\end{CD}
$$
Note that the first two rows are short exact and all vertical maps are degreewise epimorphic and weak equivalences (observe that $\operatorname{Ker}(q) = \operatorname{Ker}(p)$ is acyclic).
Thus, the map $i \colon A \to Q$ so constructed is a cofibration and the composition $Q \to \operatorname{cyl}(f) \to B$ is both a fibration and a weak equivalence.
The second factorization is a bit trickier. The main idea is to pass to the "Eckmann-Hilton dual" of the previous construction (the mapping cylinder is a homotopy push-out, its dual is a homotopy pull-back).
The hard work is to show that $f = qj$ where $q\colon D \to B$ is a fibration and $j \colon A \to D$ is a degreewise monic chain homotopy equivalence with contractible cokernel $E$.
Once this is shown, we can choose a degreewise surjective map $p \colon P \to E$ from a contractible complex $P$ with projective components and form the pull-back $Q = P \mathbin{\times_E} D$ as before to end up with the desired factorization $f = (qr)k$, where $k$ is a cofibration and a weak equivalence and $(qr)$ is a fibration:
$$
\begin{CD}
A @= A @= A \\
@V{k}V{\sim}V @V{j}V{\sim}V @VV{f}V \\
Q @>r>> D @>q>> B \\
@VVV @VVV \\
P @>p>> E
\end{CD}
$$
The above diagram is commutative and the first two columns are exact and all horizontal maps are fibrations.
Let us construct the claimed factorization of $f\colon A \to B$ into a monic chain homotopy equivalence $j \colon A \to D$ followed by a fibration $q\colon D \to B$.
In a first attempt, we factor $f \colon A \to B$ as $A \xrightarrow{i} C \xrightarrow{p} B$, where $i$ is a weak equivalence 'close to a cofibration' and $p$ is a fibration. Here $C$ is an appropriate version of the 'mapping path space' with the deficit that it might have a non-zero entry in degree $-1$. Explicitly,
let $C_n = A_n \oplus B_n \oplus B_{n+1}$ and $d_{n}^C = \begin{bmatrix} d_{n}^A & 0 & 0 \\ 0 & d_{n}^B & 0 \\ f_n & -1 & -d_{n+1}^B \end{bmatrix}$, where $C_{-1} = 0 \oplus 0 \oplus B_0$.
It is straightforward to verify that the chain map $i \colon A \to C$ with
components $\begin{bmatrix} 1 \\ f_n \\ 0 \end{bmatrix} \colon A_n \to A_n \oplus B_n \oplus B_{n+1}$ is a chain homotopy equivalence. A homotopy inverse is given by the projection $\begin{bmatrix} 1 & 0 & 0 \end{bmatrix}$ onto the first summand of $C$. Also, the chain map $i$ is degreewise monic, and its cokernel can be computed to be the contractible complex $\operatorname{cone}(1_{B[-1]})$
The lower end of the factorization of $f$ looks like this (the complexes run vertically and degrees $2,1,0,-1$ are displayed):
$$
\begin{CD}
A_2 @>{\begin{bmatrix} 1 \\ f_2 \\ 0 \end{bmatrix}}>>
A_2 \oplus B_2 \oplus B_3 @>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>> B_2 \\
@VV{d_{2}^A}V
@VV{\begin{bmatrix} d_{2}^A & 0 & 0 \\ 0 & d_{2}^B & 0 \\ f_2 & -1 & -d_{3}^B \end{bmatrix}}V
@VV{d_{2}^B}V \\
A_1 @>{i_1}>>
A_1 \oplus B_1 \oplus B_2 @>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>> B_1 \\
@VV{d_{1}^A}V
@VV{\begin{bmatrix} d_{1}^A & 0 & 0 \\ 0 & d_{1}^B & 0 \\ f_1 & -1 & -d_{2}^B \end{bmatrix}}V
@VV{d_{1}^B}V \\
A_0 @>{i_0}>>
A_0 \oplus B_0 \oplus B_1 @>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>> B_0 \\
@VVV @VV{\begin{bmatrix} f_0 & -1 & -d_{1}^B \end{bmatrix}}V
@VVV \\
0 @>>> B_0 @>>> 0 \\
\hline\\
A @>i>> C @>p>> B
\end{CD}
$$
Note that $f = pi$ and that $p$ is epimorphic in all degrees.
In order to get rid of the piece in degree $-1$, we replace $C$ by its good truncation $D = \tau_{\geq 0} C$, so
$D_n = C_n = A_n \oplus B_n \oplus B_{n+1}$ for $n \geq 1$ and $D_0 = \operatorname{Ker}(d_0\colon C_0 \to C_{-1}) = A_0 \oplus B_1$.
The inclusion $D_0 \to C_0$ is $\begin{bmatrix} 1 & 0 \\ f_0 & -d_{1}^B \\ 0 & 1 \end{bmatrix} \colon A_0 \oplus B_1 \to A_0 \oplus B_0 \oplus B_1$.
Factoring everything in sight over $D_0$, the lower end of the sequence $A \xrightarrow{j} D \xrightarrow{q} B$ turns out to be
$$
\begin{CD}
A_2 @>{\begin{bmatrix} 1 \\ f_2 \\ 0 \end{bmatrix}}>>
A_2 \oplus B_2 \oplus B_3 @>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>> B_2 \\
@VV{d_{2}^A}V
@VV{\begin{bmatrix} d_{2}^A & 0 & 0 \\ 0 & d_{2}^B & 0 \\ f_2 & -1 & -d_{3}^B \end{bmatrix}}V
@VV{d_{2}^B}V \\
A_1 @>{i_1}>>
A_1 \oplus B_1 \oplus B_2 @>{\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}}>> B_1 \\
@VV{d_{1}^A}V
@VV{\begin{bmatrix} d_{1}^A & 0 & 0 \\ f_1 & -1 & -d_{2}^B \end{bmatrix}}V
@VV{d_{1}^B}V \\
A_0 @>>{\begin{bmatrix} 1 \\ 0\end{bmatrix}}>
A_0 \oplus B_1 @>>{\begin{bmatrix} f_{0} & -d_{1}^B \end{bmatrix}}> B_0 \\
\hline \\
A @>j>> D @>q>> B
\end{CD}
$$
Observe that $q$ is a fibration (it is an epimorphism except in degree zero) and that $f = qj$.
To finish up, note that $j$ is a degreewise monic chain homotopy equivalence whose cokernel is the complex $E_n = B_{n} \oplus B_{n+1}$ in degrees $n \geq 1$ and $E_0 = B_1$, the differential is given by the matrix $\begin{bmatrix} d^{B}_n & 0 \\ -1 & -d^{B}_{n+1}\end{bmatrix}$.
From here it's now easy to write down the complex $Q$ explicitly, but I'll leave that to the interested reader.
Best Answer
Being flat in each degree is a necessary and sufficient condition.
The point is that the differentials of $B$, $X$, $Y$ and $Z$, and their grading by degree, are irrelevant.
Given a chain complex of modules $C$ with components $C_i$ ($i\in\mathbb{Z}$), let $\widetilde{C}$ denote the single module $\bigoplus_{i\in\mathbb{Z}}C_i$. Then $C\mapsto\widetilde{C}$ is a functor from complexes to modules.
Also, $\widetilde{C\otimes D}=\widetilde{C}\otimes\widetilde{D}$, and a sequence $0\to C\to D\to E\to0$ of complexes is exact if and only if $0\to \widetilde{C}\to \widetilde{D}\to \widetilde{E}\to0$ is exact.
So $0\to X\otimes B\to Y\otimes B\to Z\otimes B\to0$ is exact if and only if $0\to \widetilde{X}\otimes \widetilde{B}\to \widetilde{Y}\otimes \widetilde{B}\to \widetilde{Z}\otimes \widetilde{B}\to0$ is exact, which is the case for all short exact sequences $0\to X\to Y\to Z\to0$ if and only if $\widetilde{B}$ is flat, which is the case if and only if $B$ is flat in each degree.